Question

class 12 integration pls help​

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int_{-1}^1\tt log\bigg(\dfrac{2 + 3x}{2 - 3x} \bigg)dx$

Let assume that

$\rm :\longmapsto\:I \: = \: \displaystyle\int_{-1}^1\tt log\bigg(\dfrac{2 + 3x}{2 - 3x} \bigg)dx$

Let further assume that

$\rm :\longmapsto\:f(x) = log\bigg(\dfrac{2 + 3x}{2 - 3x} \bigg)$

Change x to - x

$\rm :\longmapsto\:f( - x) = log\bigg(\dfrac{2 + 3( - x)}{2 - 3( - x)} \bigg)$

$\rm :\longmapsto\:f( - x) = log\bigg(\dfrac{2 - 3x}{2 + 3x} \bigg)$

$\rm :\longmapsto\:f( - x) = log\bigg(\dfrac{2 + 3x}{2 - 3x} \bigg) ^{ - 1}$

$\rm :\longmapsto\:f( - x) = - \: log\bigg(\dfrac{2 + 3x}{2 - 3x} \bigg)$

$\rm :\longmapsto\:f( - x) = - f(x)$

We know,

By properties of definite integrals,

$\rm :\longmapsto\:\displaystyle\int_{-a}^a\tt f(x) \: dx = 0 \: \: if \: f( - x) = - f(x)$

Thus,

$\rm :\longmapsto\:\displaystyle\int_{-1}^1\tt log\bigg(\dfrac{2 + 3x}{2 - 3x} \bigg)dx = 0$